Solve {ccc}{1}&{2}&{-1}{1}&{m}&{0}{1}&{0}&{m}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { - 1 } \\ { 1 } & { m } & { 0 } \\ { 1 } & { 0 } & { m } \end{array} \right|$$

Evaluate

$$m\left(m-1\right)$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}1&2&-1\\1&m&0\\1&0&m\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}1&2&-1&1&2\\1&m&0&1&m\\1&0&m&1&0\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$mm=m^{2}$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$m\left(-1\right)+m\times 2=m$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$m^{2}-m$$

Subtract m from m^{2}.

$$m\left(m-1\right)$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}1&2&-1\\1&m&0\\1&0&m\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$det(\left(\begin{matrix}m&0\\0&m\end{matrix}\right))-2det(\left(\begin{matrix}1&0\\1&m\end{matrix}\right))-det(\left(\begin{matrix}1&m\\1&0\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$mm-2m-\left(-m\right)$$

Simplify.

$$m^{2}-2m-\left(-m\right)$$

Add the terms to obtain the final result.

$$m\left(m-1\right)$$

Integrate w.r.t. m

$$\frac{m^{3}}{3}-\frac{m^{2}}{2}+С$$